Question

a 1.50 gram sample of an unknown hydrocarbon was combusted to produce 4.40 gram of co₂ and 1.80 grams of h₂o. what is the empirical formula of this unknown compound?
○ ch
○ c₂h
○ ch₂
○ ch₃

Explanation:

Step1: Calculate moles of C

Molar mass of \(CO_2 = 44\space g/mol\). Moles of \(CO_2=\frac{4.40\space g}{44\space g/mol}= 0.1\space mol\). Since 1 mole \(CO_2\) has 1 mole C, moles of \(C = 0.1\space mol\). Mass of \(C=0.1\space mol\times12\space g/mol = 1.2\space g\)

Step2: Calculate moles of H

Molar mass of \(H_2O=18\space g/mol\). Moles of \(H_2O=\frac{1.80\space g}{18\space g/mol}=0.1\space mol\). Since 1 mole \(H_2O\) has 2 moles H, moles of \(H = 0.2\space mol\). Mass of \(H=0.2\space mol\times1\space g/mol = 0.2\space g\)

Step3: Find mole ratio

Mass of C + H \(=1.2 + 0.2=1.4\space g\). Since the sample is 1.50 g (hydrocarbon, so only C and H), ratio of moles \(C:H=0.1:0.2 = 1:2\)

Answer:

\(CH_2\)